I remember reading in a geology book that a seismogram is a convolution between a source signal and propagation effects. In layman's terms, what does this really mean?


2 Answers 2


In simplest terms, it simply means that:

  1. if the source signal is shifted by some amount of time Δt, but otherwise unchanged, then the seismogram will also be shifted by Δt, but otherwise unchanged; and

  2. the seismogram generated by the sum of two (or more) source signals is the sum of the seismograms that would've been generated by each of the individual source signals.

Basically, it works exactly like an ordinary echo: what you hear is the original sound, or rather several overlapping echoes of it, each reaching you by a different path, delayed and attenuated by varying amounts during its propagation.

Mathematically, this process can be represented by convolving the source signal waveform with a function, called the convolution kernel, that describes how the signal is spread out and split into multiple echoes. Specifically, the convolution kernel is northing more or less than the waveform of the echo that would, in principle, be produced by an "ideal unit impulse" source signal — that is to say, a source signal consisting of a single, infinitesimally short pulse of sound pressure carrying one unit of energy. For this reason, the convolution kernel is also known as the impulse response of the system.

The convolution process itself, in the simplest case, merely amounts to summing together several shifted and scaled copies of the impulse response. Specifically, this is the case when the input signal can be represented (or at least approximated) as a sum of several pure impulses. More generally, any input signal can be mathematically represented as a sum — or, more properly, an integral — of infinitely many tiny impulses spaced infinitely closely*, in which case the resulting seismogram is simply the integral of infinitely many infinitely small shifted copies of the impulse response. This is precisely what a convolution does.

*) Mathematicians, and perhaps some physicists, reading this will cringe at this point; while a sum of impulses can indeed approximate any function as a measure in the weak sense, it's not a good approximation is most other senses. Nonetheless, it is a rather useful and concrete way to think about convolutions.

One reason why thinking about seismograms as convolutions is useful is that, if we know the physics of seismic propagation and the geological structure of the area through which the signal propagates, we can calculate the theoretical impulse response function of the system and convolve it with a simulated source signal to obtain a synthetic seismogram. Conversely, given a real seismogram and the source signal that produced it, we can deconvolve them to obtain an approximation of the impulse response, which can then yield tomographical information about the geology.

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    $\begingroup$ Ps. I hope that's "in layman's terms" enough. ;-) $\endgroup$ Apr 19, 2014 at 9:00
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    $\begingroup$ I still find it amazing how concepts in imaging can be applied to very different domains. Perhaps that is why we covered Fast Fourier Transforms so much at univ... $\endgroup$
    – winwaed
    Apr 19, 2014 at 12:58

The reason we use convolution is because we consider the earth to be a linear, time-invariant, passive system. The output of any such system is the convolution of the input and the impulse response of the system.

"linear" means that if input x(t) produces output X(t) and input y(t) produces output Y(t), then input Ax(t)+By(t) produces output AX(t)+BY(t) [where A & B are constants, t is time]

"time-invariant" means that if input x(t) produces output X(t), then input x(t-T) will produce output X(t-T)

"passive" means that if the input starts at time T then no output is produced prior to time T.

  • $\begingroup$ Why is it that the earth be modeled as a "linear, time-invariant passive system"? $\endgroup$
    – Paul
    Jul 17, 2015 at 3:48
  • $\begingroup$ @Paul I suppose the main reason is the simplicity of such model. Moreover, as far as we can conclude, this model fits our observations well enough. There is a certain amount of investigations dedicated to nonlinear seismic effects (example), though I'm not quite familiar with them. $\endgroup$
    – antongrin
    Jul 17, 2015 at 9:09

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